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Daniel [21]
3 years ago
7

A rectangular loop with dimensions 4.20 cm by 9.50 cm carries current I. The current in the loop produces a magnetic field at th

e center of the loop that has magnitude 3.10×10−5 T and direction away from you as you view the plane of the loop. What are the magnitude and direction (clockwise or counterclockwise) of the current in the loop?
Physics
1 answer:
rusak2 [61]3 years ago
4 0

Answer:

1.63 A and in clockwise direction

Explanation:

The magnetic field due to the rectangular loop is :

$B=\frac{2 \mu_0 I}{\pi}\left(\frac{\sqrt{L^2+W^2}}{LW}\right)$

Given : W = 4.20 cm

                $=4.20 \times 10^{-2} \ m$

            L = 9.50 cm

               $= 9.50 \times 10^{-2} \ m$

            $B = 3.40 \times 10^{-5} \ T $

   Rearranging the above equation, we get

$I=\frac{B \pi LW}{2 \mu_0\sqrt{L^2+W^2}}$

$I=\frac{(3.40 \times 10^{-5}) \pi(9.50 \times 10^{-2})(4.20 \times 10^{-2})}{2(4 \pi \times 10^{-7})\sqrt{(9.50 \times 10^{-2})^2+(4.20 \times 10^{-2})^2}}$

I = 1.63 A

So the magnitude of the current in the rectangular loop is 1.63 A.

And the direction of current is clockwise.

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A water pump belong to a positive displacement pump that provides constant flow of water at fixed speed, regardless of changes in the counter pressure. The two main types of positive displacement pump are rotary pumps and reciprocating pumps.

Moreover, water pump is a reciprocating positive displacement pump that have an expanding cavity on the suction side and a decreasing cavity on the discharge side. In water pumps, the liquid flows into the pumps as the cavity on the suction side expands and then the liquid flows out of the discharge as the cavity collapses providing water in a pail.

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vaccines are used to prevent disease. they are given to patients as dead or weakened forms of a virus. which disease would most
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Flu,because it has the capacity of better fight influenza and it has no bad side effects and it can be taken easily as nasal spray or in the arm.
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3 years ago
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Light travels faster in warmer air. On a sunny day, the sun can heat a road and create a layer of hot air above it. Let's model
AysviL [449]

Answer:

Explanation:

If we assume there is a sharp boundary between the two masses of air, there will be a refraction. The refractive index of each medium will depend on the relative speeds of light.

n = c / v

If light travels faster in warmer air, it will have a lower refractive index

nh < nc

Snell's law of refraction relates angles of incidence and refracted with the indexes of refraction:

n1 * sin(θ1) = n2 * sin(θ2)

sin(θ2) = sin(θ1) * n1/n2

If blue light from the sky passing through the hot air will cross to the cold air, then

n1 = nh

n2 = nc

Then:

n1 < n2

So:

n1/n2 < 1

The refracted light will come into the cold air at angle θ2 wich will be smaller than θ1, so the light is bent upwards, creating the appearance of water in the distance, which is actually a mirror image of the sky.

6 0
4 years ago
Is space expanding within clusters of galaxies? why or why not?
GarryVolchara [31]

No, since their gravity is powerful enough to keep them together even while the universe expands as a whole. Space is not expanding within clusters of galaxies.

<h3>What is a galaxy?</h3>

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No, since their gravity is powerful enough to keep them together even while the universe expands as an entire.

Hence,space is not expanding within clusters of galaxies.

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6 0
2 years ago
If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, what is the minumum value the c
Lemur [1.5K]

Answer:

\mu_s \geq 0.27

Explanation:

The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:

f_s=\frac{mv^{2}}{R}

But we know that:

f_s\leq \mu_s N

And the normal force is given by the sum of the forces in the vertical direction:

N-mg=0 \implies N=mg

Finally, we have:

f_s=\frac{mv^{2}}{R}  \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR}  \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27

So, the minimum value for the coefficient of friction is 0.27.

4 0
3 years ago
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